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Math Help - 2nd order ODE - Euler Equation

  1. #1
    Junior Member utopiaNow's Avatar
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    2nd order ODE - Euler Equation

    Given:
    2x^2y'' + 3xy' + (2x^2 - 1)y = 0

    Question:
    Find the indicial equation and determine the two singular roots.

    I know how to solve it when the equation is of the form:
    x^2y'' + \alpha xy' + \beta y

    However here, \beta = 2x^2 - 1 and is not a constant. How am I supposed to find the indicial/characteristic equation here?

    In the solutions, they just ignore the 2x^2 factor and proceed as if the equation was 2x^2y'' + 3xy' - y = 0. I just don't understand how the 2x^2 factor can be ignored.

    Any suggestions?
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  2. #2
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    I assume you're looking for a Frobenius solution. Here's a method for finding the indicial equation when the coefficients are polynomials. Divide each term of the DE by the leading polynomial coefficient, and write the DE in the form:

    y^{\prime\prime}+\frac{p_0+p_1x+p_2x^2+\ldots}{x}y  ^{\prime}+\frac{q_0+q_1x+q_2x^2+\ldots}{x^2}=0.

    Now your indicial equation is just r(r-1)+p_0r+q_0=0.

    So in your example, we can rewrite the DE as

    y^{\prime\prime}+\frac{\frac{3}{2}}{x}y^\prime + \frac{-\frac{1}{2}+x^2}{x^2}y=0,

    so that p_0=3/2 and q_0=-1/2. Now the indicial equation is r(r-1)+\frac{3}{2}r-\frac{1}{2}=0, or (r-\frac{1}{2})(r+1)=0.

    Now you're on your way.
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